上证所加转转换

Posted 2023-02-16

【中文标题】上证所加转转换

【英文标题】：SSE addition and conversion

【发布时间】：2017-06-13 02:21:23

【问题描述】：

事情是这样的，我如何添加两个无符号字符数组并将结果存储在一个无符号短数组中，使用 SSE。谁能给我一些帮助或提示。这是我到目前为止所做的。我只是不知道错误在哪里..需要一些帮助

#include<iostream>
#include<intrin.h>
#include<windows.h>
#include<emmintrin.h>
#include<iterator>

using namespace std;

void sse_add(unsigned char * input1, unsigned char *input2, unsigned short  *output, const int N)
   

unsigned char *op3 = new unsigned char[N];
unsigned char *op4 = new unsigned char[N];

__m128i *sse_op3 = (__m128i*)op3;
__m128i *sse_op4 = (__m128i*)op4;
__m128i *sse_result = (__m128i*)output;

for (int i = 0; i < N; i = i + 16)

     __m128i src = _mm_loadu_si128((__m128i*)input1);
     __m128i zero = _mm_setzero_si128();
     __m128i higher = _mm_unpackhi_epi8(src, zero);
     __m128i lower = _mm_unpacklo_epi8(src, zero);

     _mm_storeu_si128(sse_op3, lower);
     sse_op3 = sse_op3 + 1;
     _mm_storeu_si128(sse_op3, higher);
     sse_op3 = sse_op3 + 1;
    input1 = input1 + 16;



for (int j = 0; j < N; j = j + 16)

     __m128i src1 = _mm_loadu_si128((__m128i*)input2);
     __m128i zero1 = _mm_setzero_si128();
     __m128i higher1 = _mm_unpackhi_epi8(src1, zero1);
     __m128i lower1 = _mm_unpacklo_epi8(src1, zero1);

    _mm_storeu_si128(sse_op4, lower1);
    sse_op4 = sse_op4 + 1;
    _mm_storeu_si128(sse_op4, higher1);
    sse_op4 = sse_op4 + 1;
    input2 = input2 + 16;



__m128i *sse_op3_new = (__m128i*)op3;
__m128i *sse_op4_new = (__m128i*)op4;

for (int y = 0; y < N; y = y + 8)

    *sse_result = _mm_adds_epi16(*sse_op3_new, *sse_op4_new);
    sse_result = sse_result + 1;
    sse_op3_new = sse_op3_new + 1;
    sse_op4_new = sse_op4_new + 1;




void C_add(unsigned char * input1, unsigned char *input2, unsigned short *output, int N)

for (int i = 0; i < N; i++)
    output[i] = (unsigned short)input1[i] + (unsigned short)input2[i];





int main()

int n = 1023;
unsigned char *p0 = new unsigned char[n];
unsigned char *p1 = new unsigned char[n];
unsigned short *p21 = new unsigned short[n];
unsigned short *p22 = new unsigned short[n];
for (int j = 0; j < n; j++)

    p21[j] = rand() % 256;
    p22[j] = rand() % 256;


C_add(p0, p1, p22, n);
cout << "C_add finished!" << endl;
sse_add(p0, p1, p21, n);
cout << "sse_add finished!" << endl;

for (int j = 0; j < n; j++)

    if (p21[j] != p22[j])
    
        cout << "diff!!!!!@@@@@@@" << endl;
    

//system("pause");

delete[] p0;
delete[] p1;
delete[] p21;
delete[] p22;
return 0;


 

【参考方案1】：

假设所有内容都与_Alignof(__m128i) 对齐，并且数组的大小是sizeof(__m128i) 的倍数，这样应该可以：

void addw(size_t size, uint16_t res[size], uint8_t a[size], uint8_t b[size]) 
  __m128i* r = (__m128i*) res;
  __m128i* ap = (__m128i*) a;
  __m128i* bp = (__m128i*) b;

  for (size_t i = 0 ; i < (size / sizeof(__m128i)) ; i++) 
    r[(i * 2)]     = _mm_add_epi16(_mm_cvtepu8_epi16(ap[i]), _mm_cvtepu8_epi16(bp[i]));
    r[(i * 2) + 1] = _mm_add_epi16(_mm_cvtepu8_epi16(_mm_srli_si128(ap[i], 8)), _mm_cvtepu8_epi16(_mm_srli_si128(bp[i], 8)));
  

FWIW，NEON 会更简单一些（使用vaddl_u8 和vaddl_high_u8）。

如果您正在处理未对齐的数据，您可以使用_mm_loadu_si128/_mm_storeu_si128。如果 size 不是 16 的倍数，您只需在没有 SSE 的情况下完成余数。

请注意，这可能是您的编译器可以自动执行的操作（我尚未检查）。你可能想尝试这样的事情：

#pragma omp simd
for (size_t i = 0 ; i < size ; i++) 
  res[i] = ((uint16_t) a[i]) + ((uint16_t) b[i]);

它使用 OpenMP 4，但也有 Cilk++ (#pragma simd)、clang (#pragma clang loop vectorize(enable))、gcc (#pragma GCC ivdep)，或者您可以只希望编译器足够聪明而没有编译指示提示。

【讨论】：

非常感谢，我已经发布了我的代码...运行时出现错误（断点），真的很困惑..